Question: Let $a,$ $b,$ and $t$ be real numbers such that $a + b = t.$  Find, in terms of $t,$ the minimum value of $a^2 + b^2.$
Solution: By QM-AM,
\[\sqrt{\frac{a^2 + b^2}{2}} \ge \frac{a + b}{2} = \frac{t}{2}.\]Then
\[\frac{a^2 + b^2}{2} \ge \frac{t^2}{4},\]so $a^2 + b^2 \ge \frac{t^2}{2}.$

Equality occurs when $a = b = \frac{t}{2},$ so the minimum value of $a^2 + b^2$ is $\boxed{\frac{t^2}{2}}.$